Prove that e is Irrational

The number e is an irrational number and its values lies between 2 and 3. In this article, let us learn how to prove that e is irrational.

Question

Prove that e is irrational.

Answer:

If possible, suppose that e is an irrational number. So $e=\dfrac{p}{q}$ where p and q are positive integers prime to each other.

Let us recall the exponential function as an infinite series.

e^x = $\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}$

Put x=1.

⇒ e = $\displaystyle \sum_{n=0}^\infty \dfrac{1}{n!}$

Let us write S_k = $\displaystyle \sum_{n=0}^k \dfrac{1}{n!}$.

So, e-S_k = $\displaystyle \sum_{r=1}^\infty \dfrac{1}{(k+r)!}$.

Note that

0 < e-S_k < $\dfrac{1}{(k+1)!} \left( 1+\dfrac{1}{k+1} +\dfrac{1}{(k+1)^2}+\cdots \right)$

= $\dfrac{1}{(k+1)!} \times \dfrac{1}{1-\frac{1}{k+1}}$

= $\dfrac{1}{k \cdot k!}$

That is, 0 < e-S_k < $\dfrac{1}{k \cdot k!}$ …(∗)

So we deduce that

0 < e-S_q < $\dfrac{1}{q \cdot q!}$

⇒ 0 < q! (e-S_q) < $\dfrac{1}{q}$ …(∗∗)

As e= p/q, the number $e \cdot q!$ is an integer. This implies $S_q \cdot q!$ is an integer. Therefore, from (∗∗) it follows that there must be an integer between 0 and 1, which is a contradiction. So our assumption was wrong which means e is not a rational number.

Therefore, e is an irrational number.

Real Analysis Practice Problems

FAQs